Question

A computer cooled by a fan contains eight printed circuit boards ( \(\mathrm{PCBs}\) ), each dissipating \(12 \mathrm{~W}\) of power. The height of the PCBs is \(12 \mathrm{~cm}\) and the length is \(15 \mathrm{~cm}\). The clearance between the tips of the components on the \(\mathrm{PCB}\) and the back surface of the adjacent \(\mathrm{PCB}\) is \(0.3 \mathrm{~cm}\). The cooling air is supplied by a \(10-W\) fan mounted at the inlet. If the temperature rise of air as it flows through the case of the computer is not to exceed \(10^{\circ} \mathrm{C}\), determine \((a)\) the flow rate of the air that the fan needs to deliver, \((b)\) the fraction of the temperature rise of air that is due to the heat generated by the fan and its motor, and (c) the highest allowable inlet air temperature if the surface temperature of the components is not to exceed \(70^{\circ} \mathrm{C}\) anywhere in the system. As a first approximation, assume flow is fully developed in the channel. Evaluate properties of air at a bulk mean temperature of \(25^{\circ} \mathrm{C}\). Is this a good assumption?

Short answer

The required flow rate of air for the fan is \(\dot{V} = 0.008\,\mathrm{m^3/s}\).

Step-by-step solution

A. Flow rate of the air

First, let's determine the total power dissipated by the PCBs: Total power dissipated = No. of PCBs x power dissipated by each PCB = \(8\times12\,\mathrm{W}=96\,\mathrm{W}\) Next, we need to find the total heat transfer rate needed for the air to pass over the PCBs such that the temperature rise does not exceed \(10^{\circ} \mathrm{C}\). We will use the following formula: \(Q = mc_p\Delta T\) Where \(Q\) is the heat transfer rate, \(m\) is the mass flow rate of the air, \(c_p\) is the specific heat of air at constant pressure, and \(\Delta T\) is the temperature rise of air. We know the \(\Delta T\) should not exceed \(10^{\circ} \mathrm{C}\). Solve for the mass flow rate \(m\): \(m = \frac{Q}{c_p \Delta T}\) Now, we need to find the volume flow rate of air, which is given by the following formula: \(\dot{V} = \frac{m}{\rho}\) Where \(\rho\) is the density of air. For air at \(25^{\circ} \mathrm{C}\), \(\rho \approx 1.184\,\mathrm{kg/m^3}\), and \(c_p \approx 1005\,\mathrm{J/(kg \cdot K)}\). Now, substitute the values: \(\dot{V} = \frac{96\,\mathrm{W}}{1005\,\mathrm{J/(kg \cdot K)} \times 10\,\mathrm{K} \times 1.184\,\mathrm{kg/m^3}}\) Therefore, the flow rate of air that the fan needs to deliver is: \(\dot{V} = 0.008\,\mathrm{m^3/s}\)